Vectors & 3D Geometry
Scalar and vector triple products; perpendicular vectors
MMTS_Full_Test_07
Grade 12
Question:
Let $\vec{a}$, $\vec{b}$, $\vec{c}$ have magnitudes 1, 2, 3 respectively, satisfying $|(\vec{a}\times\vec{b})\cdot\vec{c}|=6$. If $\vec{d}$ is a unit vector coplanar with $\vec{b}$ and $\vec{c}$ such that $\vec{b}\cdot\vec{d}=1$, then $|(\vec{a}\times\vec{c})\cdot\vec{d}|^2+|(\vec{a}\times\vec{c})\times\vec{d}|^2$ is
(A) 9
(B) 3
(C) 27
(D) $9/2$
Step-by-Step Solution
Key Concept: From $|(\vec{a}\times\vec{b})\cdot\vec{c}|=|\vec{a}||\vec{b}||\vec{c}|\sin\alpha\cos\beta=6=1\cdot2\cdot3$, so $\sin\alpha\cos\beta=1$, meaning $\vec{a}\perp\vec{b}$ and $\vec{b}\perp\vec{c}$, forcing $\vec{a},\vec{b},\vec{c}$ mutually perpendicular.
$\vec{a},\vec{b},\vec{c}$ mutually perpendicular. $|\vec{a}\times\vec{c}|=3$. Identity: $(\vec{u}\cdot\vec{d})^2+(\vec{u}\times\vec{d})^2=|\vec{u}|^2=9$.
Correct Answer: (A) 9